For a function $h$, we are given that $h(-2)=-5$ and $h'(-2)=-9$. What's the equation of the tangent line to the graph of $h$ at $x=-2$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $y+9=-5(x+2)$ (Choice B) B $y+5=-9(x+2)$ (Choice C) C $y+2=-5(x+9)$ (Choice D) D $y+2=-9(x+5)$
Solution: The derivative of a function at a point gives the slope of the line tangent to the function's graph at that point. Therefore, $h'(-2)$ gives the slope of the tangent line to the graph of $h$ where $x=-2$. We are given that $h'(-2)=-9$, so the slope of the tangent line is $-9$. Furthermore, we are given that $h(-2)=-5$, which means the point of intersection of the tangent line and the graph is $(-2,-5)$. To summarize, the tangent line has a slope of $-9$ and it passes through the point $(-2,-5)$. We can use the point-slope form of linear equations to find the tangent line equation: $\begin{aligned} y-y_1&=m(x-x_1) \\\\ y-(-5)&=-9(x-(-2)) \\\\ y+5&=-9(x+2) \end{aligned}$ The equation is $y+5=-9(x+2)$.